Let $e_n(x_1,x_2,x_3,\dots)$ denote the $n$-th elementary symmetric function in the infinite variables $x_1,x_2,x_3,\dots$.

Let $u$ and $v$ be the roots of $z^2-6z+1=0$.

Question. Let $x_j=\frac1{j^8}$. The following seems to be true, but can one prove or disprove? $$e_n(x_1,x_2,x_3,\dots)=\frac{4^{3n+1}\pi^{8n}(u^{2n+1}+v^{2n+1})}{(8n+4)!}.$$

For example, $e_0=1$ and $e_1=\sum_{j\geq1}\frac1{j^8}=\zeta(8)=\frac{\pi^8}{9450}$.

  • 3
    In other words, you're asking for the evaluation of the infinite product $\prod_{j=1}^{+\infty} (\frac{1}{j^8}-t)$, right? – Gro-Tsen Mar 24 '17 at 10:58
  • That's is correct. – T. Amdeberhan Mar 24 '17 at 11:03
up vote 11 down vote accepted

As Gro-Tsen suggests in the comments, we have to expand the infinite product $$f(t)=\prod_j \left(1-\frac{t^8}{j^8}\right)=\prod_{j;\,w^4=1} \left(1-\frac{\omega t^2}{j^2}\right)=\prod_{w^4=1}\frac{\sin\pi\sqrt{w}t}{\pi\sqrt{w}t},$$ we expand product of four sines as an alternating sum of cosines $$\sin a\sin b\sin c\sin d=\frac18\sum (\prod \pm)\cdot\cos(a\pm b\pm c\pm d),$$ and write Taylor series for cosines. Powers of $1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)$ appear. To be more concrete, $c_n:=e_n(x_1,\dots)$ equals $(-1)^n\times [t^{8n}]f(t)$, where $[t^n]g$ denotes a coefficient of $t^n$ in $g$. Thus $$c_n=(-1)^{n+1}i\pi^{8n}\frac1{8(8n+4)!}\sum \pm\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^{8n+4}.$$ We have $$\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^4=8i(\pm 3\pm 2\sqrt{2}),$$ that gives your formula after careful simplifications.

  • Thanks, Fedor! If you give details to reach the RHS (of the question) from your "powers of ... appear" step, then I like to accept the answer. – T. Amdeberhan Mar 24 '17 at 14:12

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